Let H be orthocenter and O circumcenter of an acuted angled triangle ABC. D and E are feets of perpendiculars from A and B on BC and AC respectively. Let OD and OE intersect BE and AD in K and L, respectively. Let X be intersection of circumcircles of HKD and HLE different than H, and M is midpoint of AB. Prove that K,L,M are collinear iff X is circumcenter of EOD.
Problem
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Tags: geometry, circumcircle, geometry unsolved
08.04.2011 23:34
I think there is a typo in problem ,I`ve found that the circumcenter of EOD always lies on OX and that its never coincide with X . Sorry if I`m an idiot !
12.04.2011 22:16
There's absolutely something wrong with the problem. K, M and L are never concurrent since angles MAL and MBK are less than 180 degrees.
12.04.2011 22:29
If you can read Serbian language or whatever it is, their official problem and solution link is http://srb.imomath.com/zadaci/2011_smo_resenja.pdf Good luck!
12.04.2011 22:37
The graphic is a little different and the problem wording is actually correct as I have seen their solution by just looking at their graphic for this problem on that web link. Sorry!
13.04.2011 06:48
The problem just does not apply to any configuration. That's all. Weird! Too busy to find out.
19.06.2011 05:44
I have a solution in Chinese,but I haven't enough time to translate it into English See here: http://bbs.eduu.com/thread-792038-1-1.html 7#~12#